要点

  • 较好的思路解析
  • \(dp[i]\)为到达\(i\)花费的最短时间，\(dis[i]-dis[j]<=lim1\)的情况其实可以省略，因为就相当于没买咖啡，绝对不优于在那之前的某店买了咖啡并发挥效用；\(lim1<dis[i]-dis[j]<lim0\)可以用一个队列维护在范围内的点，用单增的队列就可以之间取队头，注意维护时相等也要入队，因为往后滑动窗口时如果第一个被淘汰了，第二个却可能合法；\(dis[i]-dis[j]>=lim0\)，因为是从头开始在距离上都合法所以不需要队列滑动，只要用变量维护最值即可。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long ll;
typedef double db;
const int maxn = 5e5 + 5;
const db eps = 1e-8;

ll L;
int a, b, t, r, n;
int pre[maxn];
ll dis[maxn];
db dp[maxn];
db val[maxn];

int dcmp(db x) {
	if (fabs(x) < eps)	return 0;
	return x > 0 ? 1 : -1;
}

void print(int i, int cnt) {
	if (!i) {
		cout << cnt << '\n';
		return;
	}
	print(pre[i], cnt + 1);
	cout << i - 1 << " ";
}

int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);

	cin >> L >> a >> b >> t >> r >> n;
	for (int i = 1; i <= n; i++) {
		cin >> dis[i];
	}
	dis[++n] = L;

	ll lim0 = t * a + r * b;
	ll lim1 = t * a;
	int j = 1, k = 1;
	db rej = 1e18;
	int idj = 0;
	int Q[maxn], head = 1, tail = 0;

	for (int i = 1; i <= n; i++) {
		dp[i] = (db)dis[i] / a;

		while (j < i && dis[i] - dis[j] >= lim0) {
			db tmp = dp[j] - (db)dis[j] / a;
			if (dcmp(tmp - rej) < 0) {
				rej = tmp;
				idj = j;
			}
			j++;
		}
		if (idj) {
			db tmp = rej + (db)(dis[i] - lim0) / a + t + r;
			if (dcmp(dp[i] - tmp) > 0) {
				dp[i] = tmp;
				pre[i] = idj;
			}
		}

		while (k < i && dis[i] - dis[k] >= lim1) {
			val[k] = dp[k] - (db)dis[k] / b;
			while (head <= tail && dcmp(val[Q[tail]] - val[k]) > 0)	tail--;
			Q[++tail] = k++;
		}
		while (head <= tail && dis[i] - dis[Q[head]] >= lim0)	head++;
		if (head <= tail) {
			db tmp = val[Q[head]] + (db)(dis[i] - lim1) / b + t;
			if (dcmp(dp[i] - tmp) > 0) {
				dp[i] = tmp;
				pre[i] = Q[head];
			}
		}
	}
	
	print(pre[n], 0);
}